Three kids got all six of the January Math Puzzler answers correct. In fact, the three January winners were also winners in December, 2003.

The three repeat winners were brothers Eric and Phillips Hsu of Chesterfield and Leslie Hlavaty of St. Louis.

One pattern that is developing with Young Saint Louis.com's Math Puzzler competition is that entrants don't often start winning right away.

What usually happens is that it takes some time to get the rhythm of Mr. Math Puzzler's questions. Mr. Wayne Hesse is Mr. Math Puzzler. He's an eighth grade math teacher at Green Park Lutheran School in south St. Louis County.

The Math Puzzler feature has been on this website since September, 2001.

One way to develop a winning rhythm is to check out explanations of previous answers. You can do that easily. Just go to the Past Stories tab at the top of the YSL.com home page.

Pick any month after September, 2001. There will be answers to the previous month's questions.

Another way is first print out the February questions. Then go back through previous Math Puzzlers and find answers to similar questions. To check the February, 2004, questions, just click here.

After you've done your research, fill out the February entry blank, list your answers and mail in your completed entry.

These are the answers to the January Puzzlers:

January Math Puzzler answers

1. Begin with a large cube. Slice off a tiny triangular pyramid at each vertex of the cube. How many vertices does the resulting polyhedron have? How many edges does this resulting polyhedron have?

Answer: 24 vertices, 36 edges

The explanation: Every time you clip off a corner, you make three vertices where there had been one. With the original eight vertices, that means it's 3 x 8 = 24. As for the edges, each corner that is clipped off produces three edges (3 x 8 = 24). But, don't forget the original 12 edges, so 24 + 12 = 36 edges.

 

2. In one round of a TV game show, five questions are asked. The second question is worth twice as much as the first. The third question is worth three times as much as the second. The fourth question is worth four times as much as the third. The fifth question is worth five times as much as the fourth. If the fifth question is worth $12,000, what is the first question worth?

Answer: $100

The explanation: This is expressed as a formula, with F being the value of the first question. So that's

  F x 2F x 3F x 4F x 5F = 12,000

 

          120F   12,000
          ---- = ------
           120     120

 

             F = 100

 

3. The four children in the Gonzalez family have a combined age of 25 years. Maria's older brother is six times as old as she is. Her next-to-the-oldest brother is 5 years older than she is and the youngest of her brothers is twice Maria's age. How old is each of the Gonzalez children?

Answer: 2, 4, 7 and 12

The explanation: Again, this can be expressed by a simple formula, with M being the value of Maria's age:

 (6M + [M+5] + 2M + M) = 25

 

               10M + 5 = 25
                    -5   -5
               -------   --
               10M     = 20
               ---       --
                10       10

 

                     M = 2

So Maria is 2 and the other ages can be easily figured from the values of the other items in the formula.

 

4. An electrical panel has 100 switches in a row, all in the OFF position. Every second switch is turned to the ON position, and then every third switch is changed from whatever position it is in to the other position. How many switches are now in the ON position?

Answer: 51

The explanation: Mr. Math Puzzler likes questions where the answers can be arrived at by charts. And then as you figure the charts, look for patterns. This is one of them. Set up a three-column chart with Column 1 showing all switches in OFF position. Column 2 showing that every other switch is in the ON position. Column 3 will then show every third switch being changed to the opposite of what ever it was in on Column 2. As you do the chart, you'll see a pattern develop. If the number is in multiples of 2 or 3, the switch will be ON. But, if the number is a 2 times 3 or 6, the switch will be OFF. So, there are 50 times in 100 where there is a multiple of 2 and 33 times where there is a multiple of 3. But, there are 16 times in each list where there is a multiple of 6. So the answer is: 50 minus 16 = 34 and 33 minus 16 = 17 with 34 plus 17 equaling 51.

 

5. What is the least whole number that is divisible by all the whole numbers from 1 through 9?

Answer: 2,520

The explanation: This is an example of prime factorization. Look for the primes in numbers 1 through 9. One, 2 and 3 are prime numbers, 4 contains two 2s, 5 is prime, 6 has a 2 and 3, 7 is prime, 8 has three 2s and 9 has two 3s. Then, you multiple all those together: (2x3) x (3x2) x 5 x 7 = 2,520.

 

6. Find the product:

(1 - 1/2) (1 - 1/3) (1 - 1/4) (1 - 1/5) through (1 - 1/39) (1 - 1/40)

Answer: 1/40

The explanation: This is an example of cross-factoring. If you start in a series with you'd have

1   2   3   4   5   6   7   8 ....... 38   39
- x - x - x - x - x - x - x - ....... -- x --
2   3   4   5   6   7   8   9         39   40

By cross factoring, you'll end up with each denominator being a 1 and each numerator being a 1 until you get the end and the only number left is 1/40.